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katen-ka-za [31]
3 years ago
13

Please help i will give brainliest

Mathematics
1 answer:
arlik [135]3 years ago
6 0
Here’s my answers, for #10 it got cut off but I got m=9/7 and d=11.40

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A survey reported in Time magazine included the question ‘‘Do you favor a federal law requiring a day waiting period to purchase
il63 [147K]

Answer:

The  90% confidence interval for the difference between proportions is (-0.260, -0.165).

Step-by-step explanation:

<em>The question is incomplete. The complete question is:</em>

<em>"A survey reported in Time magazine included the question ‘‘Do you favor a federal law requiring a 15 day waiting period to purchase a gun?” Results from a random sample of US citizens showed that </em><em>318 of the 520 men </em><em>who were surveyed supported this proposed law while </em><em>379 of the 460 women </em><em>sampled said ‘‘yes”. Use this information to find a </em><em>90% confidence interval</em><em> for the difference in the two proportions, </em><em>pm - pw.</em><em> </em><em>Subscript pm</em><em> is the proportion of men who support the proposed law and </em><em>pw</em><em> is the proportion of women who support the proposed law. (Round answers to 3 decimal places.)"</em>

We want to calculate the bounds of a 90% confidence interval.

For a 90% CI, the critical value for z is z=1.645.

The sample of men, of size nm=-0.26 has a proportion of pm=0.612.

p_m=X_m/n_m=318/520=0.612

The sample 2, of size nw= has a proportion of pw=0.824.

p_w=X_w/n_w=379/460=0.824

The difference between proportions is (pm-pw)=-0.212.

p_d=p_m-p_w=0.612-0.824=-0.212

The pooled proportion, needed to calculate the standard error, is:

p=\dfrac{X_m+X_w}{n_m+n_w}=\dfrac{318+379}{520+460}=\dfrac{697}{980}=0.711

The estimated standard error of the difference between means is computed using the formula:

s_{pm-pw}=\sqrt{\dfrac{p(1-p)}{n_m}+\dfrac{p(1-p)}{n_w}}=\sqrt{\dfrac{0.711*0.289}{520}+\dfrac{0.711*0.289}{460}}\\\\\\s_{pm-pw}=\sqrt{0.00039+0.00045}=\sqrt{0.001}=0.029

Then, the margin of error is:

MOE=z \cdot s_{pm-pw}=1.645\cdot 0.029=0.0477

Then, the lower and upper bounds of the confidence interval are:

LL=(p_1-p_2)-z\cdot s_{p1-p2} = -0.212-0.0477=-0.260\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= -0.212+0.0477=-0.165

The  90% confidence interval for the difference between proportions is (-0.260, -0.165).

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Geometry 10th Grade. Help but only with right answers don't waste the answers
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the last one, d

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